Let $$I=\int \frac{2 x+3}{x^2+3 x} d x$$

$$\begin{array}{lc} \text { Put } & x^2+3 x=t \\ \Rightarrow & (2 x+3) d x=d t \end{array}$$

$$\begin{aligned} \therefore\quad I & =\int \frac{1}{t} d t=\log |t|+C \\ & =\log \left|\left(x^2+3 x\right)\right|+C \end{aligned}$$

Let $I=\int \frac{\left(x^2+2\right) d}{x+1} x$

$$\begin{aligned} & =\int\left(x-1+\frac{3}{x+1}\right) d x \\ & =\int(x-1) d x+3 \int \frac{1}{x+1} d x \\ & =\frac{x^2}{2}-x+3 \log |(x+1)|+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int\left(\frac{e^{6 \log x}-e^{5 \log x}}{e^{4 \log x}-e^{3 \log x}}\right) d x \\ & =\int\left(\frac{e^{\log x^6}-e^{\log x^5}}{e^{\log x^4}-e^{\log x^3}}\right) d x \quad \left[\because \operatorname{alog} b=\log b^a\right]\\ & =\int\left(\frac{x^6-x^5}{x^4-x^3}\right) d x \quad \left[\because e^{\log x}=x\right]\\ & =\int\left(\frac{x^3-x^2}{x-1}\right) d x=\int \frac{x^2(x-1)}{x-1} d x \\ & =\int x^2 d x=\frac{x^3}{3}+C \end{aligned}$$

Consider that, $$I=\int \frac{(1+\cos x)}{(x+\sin x)} d x$$

Let $$x+\sin x=t \Rightarrow(1+\cos x) d x=d t$$

$$\begin{aligned} \therefore\quad I & =\int_t^1 d t=\log |t|+C \\ & =\log |(x+\sin x)|+C \end{aligned}$$

$$\begin{aligned} \text{Let}\quad I & =\int \frac{d x}{1+\cos x}=\int \frac{d x}{1+2 \cos ^2 \frac{x}{2}-1} \\ & =\frac{1}{2} \int \frac{1}{\cos ^2 \frac{x}{2}} d x=\frac{1}{2} \int \sec ^2 \frac{x}{2} d x \\ & =\frac{1}{2} \cdot \tan \frac{x}{2} \cdot 2+C=\tan \frac{x}{2}+C \quad\left[\because \int \sec ^2 x d x=\tan x\right] \end{aligned}$$